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+ Section 8.2Estimating a Population Proportion
After this section, you should be able to…
CONSTRUCT and INTERPRET a confidence interval for a population proportion
DETERMINE the sample size required to obtain a level C confidence interval for a population proportion with a specified margin of error
DESCRIBE how the margin of error of a confidence interval changes with the sample size and the level of confidence C
Learning Objectives
+Estim
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Activity: The Beads
Your teacher has a container full of different colored beads. Your goal is to estimate the actual proportion of red beads in the container.
Form teams of 3 or 4 students.
Determine how to use a cup to get a simple random sample of beads from the container.
Each team is to collect one SRS of beads.
Determine a point estimate for the unknown population proportion.
Find a 90% confidence interval for the parameter p. Consider any conditions that are required for the methods you use.
Compare your results with the other teams in the class.
+ Conditions for Estimating p
Suppose one SRS of beads resulted in 107 red beads and 144 beads of another color. The point estimate for the unknown proportion p of red beads in the population would be
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How can we use this information to find a confidence interval for p?
ˆ p 107
2510.426
If the sample size is large enough that both npand n(1 p) are at least 10, the samplingdistribution of ˆ p is approximately Normal.
The mean of the sampling distribution of ˆ p is p.
The standard deviation of the sampling
distribution of ˆ p is ˆ p p(1 p)
n.
In practice, we do not know the value of p. If we did, we would not need toconstruct a confidence interval for it! In large samples, ˆ p will be close to p, sowe will replace p with ˆ p in checking the Normal condition.
+ Conditions for Estimating p
Check the conditions for estimating p from our sample.
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Random: The class took an SRS of 251 beads from the container.
ˆ p 107
2510.426
Normal: Both np and n(1 – p) must be greater than 10. Since we don’t know p, we check that
The counts of successes (red beads) and failures (non-red) are both ≥ 10.
nˆ p 251107
251
107 and n(1 ˆ p ) 251 1
107
251
144
Independent: Since the class sampled without replacement, they need to check the 10% condition. At least 10(251) = 2510 beads need to be in the population. The teacher reveals there are 3000 beads in the container, so the condition is satisfied.
Since all three conditions are met, it is safe to construct a confidence interval.
+ Do CYU P- 487
1. Random: Not met – this was a convenience sample. They asked the first 100 students who arrive that day and there may be a relationship between them (e.g. siblings).
Normal: this condition is met – they have n*p-hat =17 successes (those who had slept 8 hours the previous night) and n*(1-p-hat)=83 failures (those who hadn’t). Both of these are greater than 10.
Independent: this is met as long as there are at least 1000 student in the school so that the 10% condition is met.
+ P- 487: CYU : 2
Random: this is met – the inspector chose a SRS of bags.
Normal: this condition is not met. There were only 3 successes (bags with too much salt) which is less than 10. Note that there were 22 failures (bags with an appropriate amount of salt) which is greater than 10, but both values must be greater than 10 for this condition to be met.
Independent: this condition is met because we are taking a sample of less than 10% of the population (25 out of “thousands”).
+ Constructing a Confidence Interval for p
We can use the general formula from Section 8.1 to construct a confidence interval for an unknown population proportion p:
The sample proportion ̂ p is the statistic we use to estimate p.When the Independent condition is met, the standard deviation
of the sampling distibution of ̂ p is
ˆ p p(1 p)
n
Definition:
When the standard deviation of a statistic is estimated from data, the results is called the standard error(SE) of the statistic.
Since we don't know p, we replace it with the sample proportion ˆ p . This gives us the standard error (SE) of the sample proportion :
ˆ p (1 ˆ p )
n
statistic (critical value) (standard deviation of statistic)
+ Finding a Critical Value
If the Normal condition is met, we can use a Normal curve. To find a level C confidence interval, we need to catch the central area C under the standard Normal curve.
For example, to find a 95% confidence interval, we use a critical value of 2 based on the 68-95-99.7 rule. Using Table A or a calculator, we can get a more accurate critical value. Note, the critical value z* is actually 1.96 for a 95% confidence level.
statistic (critical value) (standard deviation of statistic)
In general, the central probability C under a Normal curve lies between –z* and z*. Because z* has area (1 – C)/2 to its right under the curve, we call it the upper (1 – C)/2 critical value.
This is the critical value
+ Finding a Critical Value :Use Table A to find the critical value z* for an 80% confidence interval. Assume that the Normal condition is met.
Since we want to capture the central 80% of the standard Normal distribution, we leave out 20%, or 10% in each tail. Search Table A to find the point z* with area 0.1 to its left.
So, the critical value z* for an 80% confidence interval is z* = 1.28.
The closest entry is z = – 1.28 ( By Calculator: invNorm(0.1)=-1.28).
z .07 .08 .09
– 1.3 .0853 .0838 .0823
– 1.2 .1020 .1003 .0985
– 1.1 .1210 .1190 .1170
+ Ex 31: Find z* for a 98% CI using table A or your calculator. Show your method.
invNorm( 0.99,0,1) = 2.33
Or invNorm(0.99)
Ex 32: Find z* for a 93% CI using table A or your calculator. Show your method.
invNorm( 0.965,0,1) = 1.81
Here are some common Here are some common confidence levels.confidence levels.Confidence Level
Tail area
z*
90% 0.05 1.643
95% 0.025 1.96
99% 0.005 2.576
Values z* that mark off a specified area under the Normal curve are often called CRITICAL VALUES
of the Distribution.
+ One-Sample z Interval for a Population Proportion
Once we find the critical value z*, our confidence interval for the population proportion p is
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Choose an SRS of size n from a large population that contains an unknown proportion p of successes. An approximate level C confidence interval for p is
where z* is the critical value for the standard Normal curve with area C between – z* and z*.
Use this interval only when the numbers of successes and failures in the sample are both at least 10 and the population is at least 10 times as large as the sample.
One-Sample z Interval for a Population Proportion
ˆ p z *ˆ p (1 ˆ p )
n
ˆ p z *ˆ p (1 ˆ p )
n
statistic (critical value) (standard deviation of statistic)
+ One-Sample z Interval for a Population Proportion: Calculate and interpret a 90% confidence interval for the proportion of red beads in the container. Your teacher claims 50% of the beads are red. Use your interval to comment on this claim.
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ˆ p z *ˆ p (1 ˆ p )
n
z .03 .04 .05
– 1.7 .0418 .0409 .0401
– 1.6 .0516 .0505 .0495
– 1.5 .0630 .0618 .0606 For a 90% confidence level, z* = 1.645
We checked the conditions earlier.
sample proportion = 107/251 = 0.426
statistic ± (critical value) • (standard deviation of the statistic)
0.426 1.645(0.426)(1 0.426)
251
0.426 0.051
(0.375, 0.477)
We are 90% confident that the interval from 0.375 to 0.477 captures the actual proportion of red beads in the container.
Since this interval gives a range of plausible values for p and since 0.5 is not contained in the interval, we have reason to doubt the claim.
+ Calculator commands: (WRITE ON THE YELLOW SHEET)
1- Prop Z Int
Stat , Tests, 1-Prop Z Int ( scroll down) , Enter
x=107
N=251
C-Level: .95
Calculate -Enter
+ The Four-Step Process
We can use the familiar four-step process whenever a problem asks us to construct and interpret a confidence interval.
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State: What parameter do you want to estimate, and at what confidence level?
Plan: Identify the appropriate inference method. Check conditions.
Do: If the conditions are met, perform calculations.
Conclude: Interpret your interval in the context of the problem.
Confidence Intervals: A Four-Step Process
+ Try # 40.
State: We want to estimate the actual proportion of all adults who are satisfied with the way things are going in the United States at this time at a 90% confidence level.
Plan: We should use a one-sample z-interval for p if the conditions are satisfied. Random: the adults were selected randomly. Normal: there were 256 successes (adults who were satisfied) and 769 failures (adults who were not satisfied). Both are at least 10. Independent: the sample is less than 10% of the population of all adults. The conditions are met.
Do: A 90% confidence interval :
The confidence interval, therefore, is from 0.23 to 0.27.
Conclude: We are 90% confident that the interval from 0.23 to 0.27 captures the true proportion of adults who are satisfied with the way things are going in the United States at this time.
+ Choosing the Sample SizeE
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The margin of error (ME) in the confidence interval for p is
ME z *ˆ p (1 ˆ p )
n z* is the standard Normal critical value for the level of confidence we want.
Because the margin of error involves the sample proportion ˆ p , we have to guess the latter value when choosing n. There are two ways to do this :
• Use a guess for ˆ p based on past experience or a pilot study
• Use ˆ p 0.5 as the guess. ME is largest when ̂ p 0.5
+.
To determine the sample size n that will yield a level C confidence interval for a population proportion p with a maximum margin of error ME, solve the following inequality for n:
Sample Size for Desired Margin of Error
ˆ ˆ(1 )*
ˆwhere is a guessed value for the sample proportion.
The margin of error
ˆwill always be less than or equal to if you take the guess to be 0.5.
p pz ME
np
ME p
+ Example: P- 493: Customer Satisfaction
Read the example on page 493. Determine the sample size needed to estimate p within 0.03 with 95% confidence.
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The critical value for 95% confidence is z* = 1.96.
Since the company president wants a margin of error of no more than 0.03, we need to solve the equation
1.96ˆ p (1 ˆ p )
n0.03
1.96
0.03ˆ p (1 ˆ p ) n
1.96
0.03
2
ˆ p (1 ˆ p ) n
1.96
0.03
2
(0.5)(1 0.5) n
1067.111 n
Multiply both sides by square root n and divide
both sides by 0.03.
Square both sides.
Substitute 0.5 for the sample proportion to find the largest ME
possible.
We round up to 1068 respondents to ensure the margin of error is no more than 0.03 at
95% confidence.
+ Section 8.2Estimating a Population Proportion
In this section, we learned that…
Summary
In practice, we use the sample proportion ˆ p to estimate the unknownparameter p. We therefore replace the standard deviation of ˆ p with itsstandard error when constructing a confidence interval.
The level C confidence interval for p is : ˆ p z *ˆ p (1 ˆ p )
n
Confidence intervals for a population proportion p are based on the samplingdistribution of the sample proportion ˆ p . When n is large enough that both npand n(1 p) are at least 10, the sampling distribution of p is approximatelyNormal.
+ Section 8.2Estimating a Population Proportion
In this section, we learned that…
When constructing a confidence interval, follow the familiar four-step process:
STATE: What parameter do you want to estimate, and at what confidence level? PLAN: Identify the appropriate inference method. Check conditions. DO: If the conditions are met, perform calculations. CONCLUDE: Interpret your interval in the context of the problem.
The sample size needed to obtain a confidence interval with approximate margin of error ME for a population proportion involves solving
Summary
z *ˆ p (1 ˆ p )
nME
for n, where ˆ p is a guessed value for the sample proportion, and z * is thecritical value for the level of confidence you want. If you use ̂ p 0.5 in thisformula, the margin of error of the interval will be less than or equal to ME .
+Looking Ahead…
We’ll learn how to estimate a population mean.
We’ll learn about The one-sample z interval for a population mean when σ is known The t distributions when σ is unknown Constructing a confidence interval for µ Using t procedures wisely
In the next Section…
+ 44.
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